How far have these students walked by the time the teacher's car
reaches them after their bus broke down?

At Right Angles

Stage: 3 Challenge Level:

A lot of people correctly identified that
product of the gradients of two perpendicular lines is always -1.
There were a variety of approaches to the problem:

Octavia from Melbourn Village College first
established the rule and provided examples to show it worked, and
then applied it to the lines in the problem

When you multiply the gradient of one line by the gradient of
the other line, you get negative 1, if the lines are perpendicular.
For example; look at the square. If you take the gradient of the
red line, you get -1. The gradient of the blue line is 1. 1 lot of
-1 is -1.

All perpendicular lines follow this rule; it always equals minus
one. You can show this in a table.

Line 1

Line 2

Product

$1$

$-1$

$-1$

$\frac{3}{2}$

$\frac{-2}{3}$

-1

$3$

$\frac{-1}{3}$

$-1$

The formula is $L_1=L_2\times(-1)$ e.g. take the first example.
$-1\times(-1)=1$.

Does this work with
the second example, Octavia?

The two sets of lines in 'At Right Angles' are not
perpendicular, because when the gradients of the pairs of lines are
multiplied together they do not equal -1.

Take the second example. One line is -1.333, and the other is
0.666, which multiplies to make -0.888.

The first pair of lines is perpendicular $(-2\times 0.5=-1)$

The second pair of lines is not perpendicular

The third pair of lines is perpendicular $(1.5 \times
\frac{-2}{3}=-1)$

Nadia from Melbourn Village College described
the relationship in a slightly different but equally valid way

For every pair of perpendicular lines one line's gradient will
be $\frac{a}{b}$ and the other lines gradient will be
$\frac{-b}{a}$.

Beth from Melbourn Village College provided
the following very extensive solution to all the different parts of
the problem

Is there a relationship between the gradients of perpendicular
lines?

Yes, because if the two lines are perpendicular then one of the
gradients has to be positive and the other negative. The line with
the negative gradient is the inverted fraction of the positive
gradient line. For example, positive gradient $\frac{1}{2}$
negative gradient -2 (also written as $\frac{-2}{1})$. In addition,
each angle between the lines where they meet is a right-angle.
However when the lines overlay the $x$- and $y$- axis neither of
these lines are either positive or negative but they are still
perpendicular.

Can you use your relationship to explain why the two sets of
lines above are not perpendicular?

For the first set of lines:

The line with the positive gradient has a gradient of
$\frac{3}{7}$. The line with the negative gradient has a gradient
of $\frac{-5}{2}$. The second gradient is not the first one
inverted, so therefore the two lines are not perpendicular.

For the second set of lines:

The line with the positive gradient has a gradient of
$\frac{2}{3}$. The line with the negative gradient has a gradient
of $\frac{-4}{3}$. The second gradient is not the first one
inverted, so therefore the two lines are not perpendicular.

Decide whether the two lines are perpendicular or not, and
explain how you know.

For the first set of lines:

In this case the line with the positive gradient has a gradient
of $\frac{1}{2}$. The line with the negative gradient has a
gradient of $\frac{-2}{1}$. The second gradient is thr first one
inverted, so therefore the two lines are perpendicular.

For the second set of lines:

Both lines have negative gradients, one line has the gradient of
$\frac{-1}{4}$ and the other $\frac{-5}{2}$, so they are not
perpendicular because the second gradient is not the first one
inverted and also if both lines are negative then they can't be
perpendicular.

For the third set of lines:

The line with the positive gradient has a gradient of
$\frac{11}{7}$. The line with the negative gradient has a gradient
of $\frac{-2}{3}$. The second gradient is not the first one
inverted, so therefore the two lines are not perpendicular.

Can you decide without plotting the points?

Yes, there is an equation which can help you to find out the
gradient. To find the gradient of a line you do $y$ over $x$, and
because there are two points you need to find the difference
between the $y$-coordinates and the $x$-coordinates. The equation
is $\frac{y_1-y_2}{x_1-x_2}$ this equation will give you the
gradient of the line which you can then work out from if the lines
are perpendicular or not. Once you have the two gradients if when
you multiply them together the answer is -1, then the two lines are
perpendicular.

Steve (no school listed), described an
interesting method for demonstrating why the products of gradients
must be -1

Draw the same line on two grids. Then rotate one by 90 degrees.
This gives me two lines which are at right angles. Since the grids
are square grids I know that the grids will lie on top of each
other after this, so I can put the second line back on the first
grid to get two perpendicular lines.

If the first line goes across $+x$ and up $+y$ from one end to
the next then the second must obviously go across $-y$ and up
$+x$.

The gradient of the first line will be $\frac{+y}{x}$ whereas
the gradient of the second line will be $\frac{-x}{y}$. So the
gradient of the first line is -1/(gradient of the second line).

Ashley from Durrington High School came up
with a memorable summary of the relationship

I find that it is very easy to remember with these words;
negative reciprocal. If you have $y=3x$, your perpendicular line
will be $y=\frac{-1}{3}x$ and if you have $y=\frac{-2}{3}x$, then
your perpendicular line is $y=\frac{3}{2}x$.

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